Can anyone solve x^2 +37=2x?

Question

thank you all

Answer 1

this cannot be solved with real numbers... but it is possible with imaginaries...

set = 0

x^2-2x+37

quick quadratic formula...

1+or-(-36)^(1/2)

gives you

1(+or-)6i because you cannot take a square root of a negative number

Answer 2

Quadratic Equation Method

x^2 + 37 = 2x    (Subtract 2x from both sides)

x^2 - 2x + 37 = 0

a = 1, b = - 2, c = 37

1) x= [ -b + √(b^2 - 4ac)]/(2a)

2) x= [ -b - √(b^2 - 4ac)]/(2a)

1) x = [2 + √((-2)^2 - (4 * 1 * 37))]/(2 * 1)

       = [2 + √(4 - 148)]/2

       = [2 + √(-1 * 144)]/2

       = [2 + 12 * √(-1)]/2

       = [2 + 12 i]/2

       = 1 + 6i

2) x = [2 - √((-2)^2 - (4 * 1 * 37))]/(2 * 1)

        = [2 - √(4 - 148)]/2

        = [2 - √(-1 * 144)]/2

        = [2 - 12 * √(-1)]/2

        = [2 - 12 i]/2      

        = 1 - 6i

Answer 3

the steps:

x^2+37=2x

x^2-2x+37=0

(x-1)(x-1)=0

x=1